__How to write:__

### Z = √3 + i .... In trigonometric form

#### Seing as this is a complex number the formula z = r(cosΘ + isinΘ) may come in handy here

#### With r being the modulus of z using the ARGAND diagram will help you to obtain the vaule: r= √(√ 3)² + √(1)²

#### with the imaginary part being i going up the y axis and root 3 being the real going along the x axis.

####

#### Θ in the formula is the argument of z so using the ARGAND diagram again the value that should be obtained is: arg(z) = Θ = tan⋅1(im/re) = tan⋅1(1/√3)

#### = ∏/6 radians or 30˚

#### Therefore z = 2(cos∏/6 + isin∏/6)

But we want zⁿ so we need the fact that Zⁿ = rⁿ(cosnΘ + isinnΘ)

#### Now needs to be in radians so zⁿ = 2ⁿ(cosn∏/6 + isinn∏/6)

But if we wanted this to be real for example we would have to try and make the imaginary part = 0 so to do this n=6.

## Comments (4)

## Drew H said

at 11:38 pm on Nov 17, 2008

ohhh the fun..

FP1 - Mathematical induction to show....

(r=1 on bottom) Σ (n on top) r x 2(Power r) = 2(Power n - 1) + 2

(note: i will not be using r=1 throughout though you should)

Step 1 ----->> show true for n =1

LHS when n=1 is Σ (1 on top) r x 2(Power r) = 1 x 2(Power 1)

RHS when n=1 is 2(power 1+1)x(1-1) + 2... = 0 + 2... = 2

∴ RHS = LHS when n=1

step 2 and 3 can vary so be sure to practise all types of these questions..

Step 2 ------>> Assume true for n = k

Σ (K on top) r x 2(Power r) = 2(Power k+1) x (k-1) + 2

Step 3 ---->> using the assumption to show that it it is also true for n = k+1

LHS when n = k+1 .. ∑(k+1 on top) r x 2(Power r) which is.. ∑(k on top) + (k+1)th term

Now Using the assumption from the statement made the line above..

∑(k+1 on top) r x 2(Power r) ≡ ∑(k on top) r x 2(Power r) + (k+1) x 2(Power k+1)

≡ (( 2(Power k+1) x (k-1) + 2 )) + (( (k+1) x 2(Power k+1) ))

^ ^

^ ^

(Assumption made earlier with n=k) (( k+1)th term)

Now from this you may see it more visible writing down the powers for yourself but there is a common factor of 2(Power k+1)

so..

= 2(Power k+1)[ ( k-1) +( k+1) ] + 2

= 2(Power k+1) x [ 2k ] + 2

now taking the 2 out

= 2(Power k+2) x (k) + 2

now when we look at RHS when n = k+1

2(Power (k+1) + 1) x ( (k+1) - 1 ) + 2

2(Power k+2) x (k) + 2

Which is exactly the same as LHS when n = k+1

So this means also true for n = k so implies true for n =k+1,

therefore It will be true for all other positive integers by induction.

## Drew H said

at 11:42 pm on Nov 17, 2008

wow that was a mess so i decided to put it on comment, its not my fault i didnt realise the question wouldve been so long..

## Steph Richards said

at 12:22 am on Nov 18, 2008

Awsome Drew!!!! Nice to see you getting involved. Have a look at your brothers work on the year 12 class page!

## Drew H said

at 12:34 am on Nov 18, 2008

why ty =P

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